#30
We need to find the value of
[tex]cot(\frac{-8\pi}{3})[/tex]
1. We will add 2pi to it until change it from negative to positive, then look for its quadrant
[tex]\begin{gathered} \frac{-8\pi}{3}+2\pi=\frac{-8\pi}{3}+\frac{6\pi}{3}=\frac{-2\pi}{3} \\ \\ \frac{-2\pi}{3}+2\pi=\frac{-2\pi}{3}+\frac{6\pi}{3}=\frac{4\pi}{3} \end{gathered}[/tex]
The angle 4pi/3 lies on the 3rd quadrant
Since the angle in the 3rd quadrant has the form
[tex]\pi+\theta[/tex]
Where theta is an acute angle, then
[tex]\begin{gathered} \pi+\theta=\frac{4}{3}\pi \\ \theta=\frac{4\pi}{3}-\pi \\ \theta=\frac{4\pi}{3}-\frac{3\pi}{3} \\ \theta=\frac{\pi}{3} \end{gathered}[/tex]
Then we will find the value of tan(pi/3), then reciprocal it to find cot
[tex]tan\frac{\pi}{3}=\sqrt{3}[/tex]
Reciprocal it
[tex]cot(\frac{\pi}{3})=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}[/tex]
In the 3rd quadrant tan and cot are positive values, then
The value of
[tex]\begin{gathered} cot(\pi+\frac{\pi}{3})=\frac{\sqrt{3}}{3} \\ \\ cot(\frac{4\pi}{3})=\frac{\sqrt{3}}{3} \\ \\ cot(\frac{-8\pi}{3})=\frac{\sqrt{3}}{3} \end{gathered}[/tex]
The answer is
[tex]\frac{\sqrt{3}}{3}[/tex]
